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受欢迎的三角函数 >

4cos(2θ)+16cos(θ)+4=9cos(θ)

解答

4 cos (2 θ) + 16 cos (θ) + 4 = 9 cos (θ)

解答

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = 2.63623 \dots + 2 πn, θ = - 2.63623 \dots + 2 πn

+1

度数

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 151.04497 \dots^{\circ} + 36 0^{\circ} n, θ = - 151.04497 \dots^{\circ} + 36 0^{\circ} n

求解步骤

4 cos (2 θ) + 16 cos (θ) + 4 = 9 cos (θ)

两边减去

9 cos (θ)

4 cos (2 θ) + 7 cos (θ) + 4 = 0

使用三角恒等式改写

4 + 4 cos (2 θ) + 7 cos (θ)

使用倍角公式:

cos (2 x) = 2 cos^{2} (x) - 1

= 4 + 4 (2 cos^{2} (θ) - 1) + 7 cos (θ)

化简

4 + 4 (2 cos^{2} (θ) - 1) + 7 cos (θ) : 8 cos^{2} (θ) + 7 cos (θ)

4 + 4 (2 cos^{2} (θ) - 1) + 7 cos (θ)

乘开

4 (2 cos^{2} (θ) - 1) : 8 cos^{2} (θ) - 4

4 (2 cos^{2} (θ) - 1)

使用分配律:

a (b - c) = ab - a c

a = 4, b = 2 cos^{2} (θ), c = 1

= 4 \cdot 2 cos^{2} (θ) - 4 \cdot 1

化简

4 \cdot 2 cos^{2} (θ) - 4 \cdot 1 : 8 cos^{2} (θ) - 4

4 \cdot 2 cos^{2} (θ) - 4 \cdot 1

数字相乘:

4 \cdot 2 = 8

= 8 cos^{2} (θ) - 4 \cdot 1

数字相乘:

4 \cdot 1 = 4

= 8 cos^{2} (θ) - 4

= 8 cos^{2} (θ) - 4

= 4 + 8 cos^{2} (θ) - 4 + 7 cos (θ)

化简

4 + 8 cos^{2} (θ) - 4 + 7 cos (θ) : 8 cos^{2} (θ) + 7 cos (θ)

4 + 8 cos^{2} (θ) - 4 + 7 cos (θ)

对同类项分组

= 8 cos^{2} (θ) + 7 cos (θ) + 4 - 4

4 - 4 = 0

= 8 cos^{2} (θ) + 7 cos (θ)

= 8 cos^{2} (θ) + 7 cos (θ)

= 8 cos^{2} (θ) + 7 cos (θ)

7 cos (θ) + 8 cos^{2} (θ) = 0

用替代法求解

7 cos (θ) + 8 cos^{2} (θ) = 0

令

: cos (θ) = u

7 u + 8 u^{2} = 0

7 u + 8 u^{2} = 0 : u = 0, u = - \frac{7}{8}

7 u + 8 u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

8 u^{2} + 7 u = 0

使用求根公式求解

8 u^{2} + 7 u = 0

二次方程求根公式:

若

a = 8, b = 7, c = 0

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 8 \cdot 0}{2 \cdot 8}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 8 \cdot 0}{2 \cdot 8}

7^{2} - 4 \cdot 8 \cdot 0 = 7

7^{2} - 4 \cdot 8 \cdot 0

使用法则

0 \cdot a = 0

= 7^{2} - 0

7^{2} - 0 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 7

u_{1, 2} = \frac{- 7 \pm 7}{2 \cdot 8}

将解分隔开

u_{1} = \frac{- 7 + 7}{2 \cdot 8}, u_{2} = \frac{- 7 - 7}{2 \cdot 8}

u = \frac{- 7 + 7}{2 \cdot 8} : 0

\frac{- 7 + 7}{2 \cdot 8}

数字相加/相减:

- 7 + 7 = 0

= \frac{0}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{0}{16}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 7 - 7}{2 \cdot 8} : - \frac{7}{8}

\frac{- 7 - 7}{2 \cdot 8}

数字相减:

- 7 - 7 = - 14

= \frac{- 14}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 14}{16}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{14}{16}

约分:

2

= - \frac{7}{8}

二次方程组的解是:

u = 0, u = - \frac{7}{8}

u = cos (θ)

代回

cos (θ) = 0, cos (θ) = - \frac{7}{8}

cos (θ) = 0, cos (θ) = - \frac{7}{8}

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

cos (θ) = 0

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = - \frac{7}{8} : θ = arccos (- \frac{7}{8}) + 2 πn, θ = - arccos (- \frac{7}{8}) + 2 πn

cos (θ) = - \frac{7}{8}

使用反三角函数性质

cos (θ) = - \frac{7}{8}

cos (θ) = - \frac{7}{8}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

θ = arccos (- \frac{7}{8}) + 2 πn, θ = - arccos (- \frac{7}{8}) + 2 πn

θ = arccos (- \frac{7}{8}) + 2 πn, θ = - arccos (- \frac{7}{8}) + 2 πn

合并所有解

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = arccos (- \frac{7}{8}) + 2 πn, θ = - arccos (- \frac{7}{8}) + 2 πn

以小数形式表示解

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = 2.63623 \dots + 2 πn, θ = - 2.63623 \dots + 2 πn

作图

流行的例子

sin(θ)=(7pi)/3

sin (θ) = \frac{7 π}{3}

tan(x)*cos^2(x)-tan(x)=0

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sin(x+pi)-sin(x)+sqrt(2)=0

sin (x + π) - sin (x) + 2 = 0

3 tan (x) = - 3

2(sin(x))^2-5cos(x)+1=0

2 (sin (x))^{2} - 5 cos (x) + 1 = 0